Last visit was: 23 Apr 2026, 14:18 It is currently 23 Apr 2026, 14:18
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
smartass666
Joined: 11 Feb 2012
Last visit: 02 Apr 2013
Posts: 10
Own Kudos:
492
 [89]
Given Kudos: 11
Posts: 10
Kudos: 492
 [89]
5
Kudos
Add Kudos
84
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
mikemcgarry
User avatar
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Last visit: 06 Aug 2018
Posts: 4,474
Own Kudos:
30,880
 [42]
Given Kudos: 130
Expert
Expert reply
Posts: 4,474
Kudos: 30,880
 [42]
33
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
User avatar
monsama
Joined: 20 Sep 2012
Last visit: 25 May 2024
Posts: 43
Own Kudos:
181
 [12]
Given Kudos: 34
Concentration: Finance, Entrepreneurship
GMAT 1: 750 Q50 V40
GPA: 3.37
WE:Analyst (Consulting)
GMAT 1: 750 Q50 V40
Posts: 43
Kudos: 181
 [12]
7
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
User avatar
GGMU
User avatar
Current Student
Joined: 04 Feb 2014
Last visit: 11 Aug 2018
Posts: 181
Own Kudos:
Given Kudos: 164
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3
WE:Project Management (Manufacturing)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Three is the largest number that can be divided evenly into 27 and the positive integer x
So x contains 3

10 is the largest number that can be divided evenly into both 100 and x
So x contains 2 and 5

So x = 2.3.5.n

largest possible number that could be divided into x and 2100
2100= 3*2*2*5*5*7
x = 2.3.5.n

So largest no should be 300=3*5*5*2*2 as it contains 2,3 and 5

What am I doing wrong???
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,777
Own Kudos:
13,047
 [4]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,777
Kudos: 13,047
 [4]
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi anurag16,

This question can be dealt with in a couple of different ways, but it ultimately comes down to Prime Factorization.

We're told that the largest number that divides into 27 and X is 3.

Since 27 = (3)(3)(3), that means that X's prime factorization can contain JUST ONE 3 (although it can contain other prime factors). If it contained more than one 3, then "3" would NOT be the largest number that would divide into 27 and X.

So X could be 3, 6, 12, 15, 30, etc.

Next, we're told that the largest number that divides into 100 and X is 10.

Since 100 = (2)(2)(5)(5), that means that X's prime factorization can contain JUST ONE 2 AND JUST ONE 5 (although it can contain other prime factors).

So we know that X's prime factorization consists of ONE 2, ONE 3 and ONE 5 and possibly some other primes.

We're asked for the LARGEST number that could divide into X and 2100.

2100 = (3)(7)(2)(2)(5)(5). Using what we know about X, the LARGEST number that could divide X AND 2100 would be (2)(3)(5)(7) = 210.

Final Answer:
GMAT assassins aren't born, they're made,
Rich
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,396
 [10]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
 [10]
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
smartass666
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700

This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute.
First, let's quickly analyse what we are given:
3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3.
10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5.
So x has exactly one 3, one 2 and one 5.

Which of the following is the largest possible number that could be divided into x and 2100?
We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5.
2100 = 2^2 * 3 * 5^2 * 7
It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x.
What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210
So 210 is the largest possible number that could be divided into x and 2100.

Answer (C)

Check out this post: https://anaprep.com/number-properties-r ... e-factors/
User avatar
shasadou
Joined: 12 Aug 2015
Last visit: 24 Nov 2022
Posts: 219
Own Kudos:
Given Kudos: 1,475
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE:Management Consulting (Consulting)
GMAT 3: 600 Q47 V27
Posts: 219
Kudos: 3,179
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This question is tricky because of the convoluted language. So x has one 3, one 5 and one 2 as prime factors, and also could have some other primes as no contraints have been set: x = 3*5*2*n. That is 30, 30*7, 30*11, 30*17...

2100 = 7*3*5*2*5*2. These two numbers obviously share such primes as 3*5*2. Since x cannot have more 3s, 2s, and 5s as prime factors we can borrow one 7 from 2100 to construct our max possible common factor of the 2100 and x = 30*7=210.
User avatar
Hoozan
Joined: 28 Sep 2018
Last visit: 30 Dec 2025
Posts: 646
Own Kudos:
Given Kudos: 248
GMAT 1: 660 Q48 V33 (Online)
GMAT 2: 700 Q49 V37
Products:
GMAT 2: 700 Q49 V37
Posts: 646
Kudos: 735
Kudos
Add Kudos
Bookmarks
Bookmark this Post
KarishmaB
smartass666
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700

This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute.
First, let's quickly analyse what we are given:
3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3.
10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5.
So x has exactly one 3, one 2 and one 5.

Which of the following is the largest possible number that could be divided into x and 2100?
We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5.
2100 = 2^2 * 3 * 5^2 * 7
It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x.
What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210
So 210 is the largest possible number that could be divided into x and 2100.

Answer (C)

KarishmaB can we say that because of the word COULD our answer is (C) and not (A)? Had the question been "which of the following MUST divide into" then (A) would be our answer since we are 100% sure that x will have one 3 and one 10 BUT not sure if it will have a 7,
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,396
 [1]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hoozan
KarishmaB
smartass666
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700

This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute.
First, let's quickly analyse what we are given:
3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3.
10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5.
So x has exactly one 3, one 2 and one 5.

Which of the following is the largest possible number that could be divided into x and 2100?
We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5.
2100 = 2^2 * 3 * 5^2 * 7
It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x.
What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210
So 210 is the largest possible number that could be divided into x and 2100.

Answer (C)

KarishmaB can we say that because of the word COULD our answer is (C) and not (A)? Had the question been "which of the following MUST divide into" then (A) would be our answer since we are 100% sure that x will have one 3 and one 10 BUT not sure if it will have a 7,

Yes Hoozan
Your analysis is correct.

Check out this post on factors too: https://anaprep.com/number-properties-r ... e-factors/

and these videos on Factors and Factorisation:
https://www.youtube.com/watch?v=DxIH8rjhpKY
https://www.youtube.com/watch?v=Kd-4cH4cqHw
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,962
Own Kudos:
Posts: 38,962
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109785 posts
Tuck School Moderator
853 posts